# Difficult Recurrence

I am trying to solve a Sangaku problem.

The blue circles have radii one. The goal is to find the total area of all the other circles (the three sequences of circles repeat ad infinitum).

I have almost solved the problem. I have found the area of the red circle, and the total area of all circles touching the horizontal axis. I even have a proposed recurrence for the radius of the circles going up the center. Here is my logic:

Take any circle on the vertical stack. Draw a right triangle with one vertex at its center, one at the center of the (left) blue circle, and one vertex below the blue circle's center level with the other vertex. One leg of this triangle has length one, and the hypotenuse is $1 + r$, where $r$ is the radius of the circle in question. The other leg has length one minus the sum of $r$ and the diameters of all circles below. There is just one unknown and these quantities are uniquely related by the Pythagorean theorem.

So we have a recurrence. $R(n)$ denotes the radius of the $n^{th}$ circle on the vertical stack, with $n=1$ the red circle. After manipulating the expression from the Pythagorean theorem, we get \begin{align*} R(1) &= 1/4 \\ R(n) &= \dfrac{\left(1 - 2\sum_{k = 1}^{n-1} R(k)\right)^2}{4\left(1 - \sum_{k = 1}^{n-1} R(k)\right)} \end{align*}

This is where I am stuck. I am looking for any sort of help in solving for a closed form of $R$. Thanks

(For those interested, here is my work on the first part of the problem. If you spot a mistake here, feel free to point it out, but my main question is how to solve the above recurrence.)

# Revival of Sangaku

Take any three circles that are $(1)$ all tangent to the horizontal axis and $(2)$ touching one another. The radius of the middle one can be given in terms of the radii of the two outer circles by $$\frac{1}{\sqrt{r_{mid}}} = \frac{1}{\sqrt{r_1}} + \frac{1}{\sqrt{r_2}}$$

Let's see why the relationship holds. Let $(x_{mid}, y_{mid})$ denote the center of the middle circle. Let, let $(x_1, y_1)$ and $(x_2, y_2)$ be the centers of the outer circles, with radii equal to $r_1$ and $r_2$.

Draw three right triangles:

• $\triangle_1$ with vertices $(x_1, y_1)$, $(x_{mid}, y_{mid})$, and $(x_1, y_{mid})$,
• $\triangle_2$ with vertices $(x_2, y_2)$, $(x_{mid}, y_{mid})$, and $(x_2, y_{mid})$, and
• $\triangle_3$ with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_1, y_2)$.

The hypotenuse of each triangle is the sum of the radii of the two circles involved. Furthermore, one leg equals the difference of the larger radius from the smaller. Thus, the length of the third leg equals twice the square root of the product of the radii involved. (Draw a picture here, use Pythagorean's Theorem)

Furthermore, the sum of the lengths of the third leg of $\triangle_1$ and $\triangle_2$ equal the length of the the third leg of $\triangle_3$.

Formally, \begin{align*} (r_1 + r_{mid})^2 &= leg(\triangle_1)^2 + (r_1 - r_{mid})^2 \\ (r_2 + r_{mid})^2 &= leg(\triangle_2)^2 + (r_2 - r_{mid})^2 \\ (r_1 + r_2)^2 &= leg(\triangle_3)^2 + (r_1 - r_2)^2 \\ leg(\triangle_1) + leg(\triangle_2) &= leg(\triangle_3) \\ &\implies \\ \sqrt{(r_1 + r_{mid})^2 - (r_1 - r_{mid})^2} &+ \sqrt{(r_2 + r_{mid})^2 - (r_2 - r_{mid})^2} \\ &= \sqrt{(r_1 + r_2)^2 - (r_1 - r_2)^2} \\ &\implies \\ 2\sqrt{r_1r_{mid}} + 2\sqrt{r_2r_{mid}} &= 2\sqrt{r_1r_2} \\ &\implies \\ \frac{1}{\sqrt{r_1}} + \frac{1}{\sqrt{r_2}} &= \frac{1}{\sqrt{r_{mid}}} \end{align*}

Now let's focus on the sequence of circles descending to the left. Let $R : \{1, 2, 3, ... \} \rightarrow \mathbb{R}$ define the radius of the $n^{th}$ circle, where we start with the left circle of radius one, then move to the red circle, then left to the green circle, then to the orange, etc.

\begin{align*} R(1) &= 1 \\ \frac{1}{\sqrt{R(n)}} &= \dfrac{1}{\sqrt{R(n-1)}} + 1 \end{align*} Notice that we can make the following substitution: $$T(n) = \frac{1}{\sqrt{R(n)}}$$ Then $T(n) = T(n-1) + 1$, with $T(1) = 1$, so $T(n) = n$. Solving for $R(n)$ gives us

$$R(n) = \dfrac{1}{n^2}$$

Thus each circle has area $\pi n^{-4}$.

So the total area of the nested circles touching the horizontal line is $$\pi\left(2^{-4} + 2\sum_{k=3}^{\infty} k^{-4}\right)$$

Now, you can use rigorous elementary methods, the Riemann zeta function, or WolframAlpha to compute $$\sum_{k=1}^{\infty} k^{-4} = \frac{\pi^4}{90}$$ So, $$\sum_{k=3}^{\infty} k^{-4} = \frac{\pi^4}{90} - \frac{17}{16}$$ and so the total area of the circles touching the horizontal axis is $$\pi\left(\frac{\pi^4}{45} - \frac{33}{16}\right)$$

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Are you aware of the Descartes circle theorem (mathworld.wolfram.com/DescartesCircleTheorem.html)? You can treat the real line as a circle with an infinite radius, i.e. a circle with zero curvature. Anyway, I am quite sure to have already seen such series somewhere next to Apollonian gaskets :) – Jack D'Aurizio Jul 26 '14 at 2:52
In this page - en.wikipedia.org/wiki/Apollonian_gasket - there is a line ([...] it follows that we may move from one quadruple of curvatures to another by Vieta jumping, just as we do when finding a new Markov number) suggesting the approach needed to solve your recurrence. – Jack D'Aurizio Jul 26 '14 at 2:59
For istance, I get that the curvatures $k_n=\frac{1}{R(n)}$ are $k_1=4,k_2=12,k_3=24,k_4=40,\ldots$ and they satisfy the recurrence relation $k_{n+1}=2k_n+4-k_{n-1}$. – Jack D'Aurizio Jul 26 '14 at 3:08
Hence we have $R(n)=\frac{1}{2n(n+1)}$, easy and clean. – Jack D'Aurizio Jul 26 '14 at 3:11
I'll take this to my heart, I gave you a simple proof for $R(n)$ avoiding radicals, I gave you the formula for the missing area, but you accepted the other answer. Shame on me. :D – Jack D'Aurizio Jul 26 '14 at 3:31

I believe your recurrence should have the form $$R(n) = \frac{\left(1 - 2 \sum_{k=1}^{n-1} R(k)\right)^2}{4\left(1 - \sum_{k=1}^{\color{red}{n-1}} R(k)\right)}.$$ To solve this, let $S(n) = \sum_{k=1}^n R(k)$, and solve the above expression for $S(n-1)$: $$S(n-1) = \frac{1}{2}\left(1 - R(n) - \sqrt{R(n)(2+R(n))}\right),$$ where we take the smaller root since we require $S(0) = 0$ for $R(1) = \frac{1}{4}$. Then we observe that $S(n) - S(n-1) = R(n)$, or $$2R(n) = R(n) - R(n+1) + \sqrt{R(n)(2+R(n))} - \sqrt{R(n+1)(2+R(n+1))}.$$ Now solving this recurrence for $R(n+1)$ yields $$R(n+1) = \frac{R(n)}{1 + 2R(n) + 2 \sqrt{R(n)(2+R(n))}}.$$ It is not difficult to show that if $R(1) = \frac{1}{4}$, then we can solve this recurrence explicitly to get $$R(n) = \frac{1}{2n(n+1)}.$$

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Absolutely, the upper bound $n$ in the sum was an honest typo. Thank you for the help! I should have thought of solving for the value of $S$ instead... At least I can add that to my repertoire now. – A.E Jul 26 '14 at 3:33

As stated in the comments, due to the Descartes circle theorem four mutually tangent circles satisfy the identity: $$2(\kappa_1^2+\kappa_2^2+\kappa_3^2+\kappa_4^2)=(\kappa_1+\kappa_2+\kappa_3+\kappa_4)^2$$ where $\kappa_i$ is the curvature of the circle $\Gamma_i$, i.e. the reciprocal of the radius.

By setting $\kappa_n=\frac{1}{R(n)}$, Vieta jumping gives: $$\kappa_0=0,\kappa_1=4,\quad\kappa_{n+1} = 2\kappa_n -\kappa_{n-1}+4,$$ hence: $$R(n)=\frac{1}{2n(n+1)}$$ and: $$\sum_{n=1}^{+\infty}R(n)^2 = \frac{\pi^2-9}{12},$$ since: $$\frac{1}{n^2(n+1)^2}=\left(\frac{1}{n}-\frac{1}{n+1}\right)^2=\frac{1}{n^2}+\frac{1}{(n+1)^2}-\left(\frac{2}{n}-\frac{2}{n+1}\right),$$ so $\zeta(2)=\frac{\pi^2}{6}$ and a new rational number make their appearance.

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Seriously, thank you! I guess I am just a wee bit prideful and wanted to see my path to the solution fulfilled. I wish I could accept both answers but heropup's was exactly what I asked for. – A.E Jul 26 '14 at 3:36
@Jack D'Aurizio Can you explain how you came up with the recurrence κn+1=2κn−κn−1+4? I have read about Vieta Jumping on Wikipedia and it only talks about using it for proving by contradiction – Rushil Jul 5 '15 at 13:46

A useful technique for dealing with circle packing problems of this type is circle inversion. Since inversion preserves tangency and sends every line and circle to another line or circle (with circles passing through the center of the circle of inversion becoming lines), if we invert with respect to a circle of radius $1$ centered at one of the corners of the "triangle" in the set up, we are left analyzing a circle packing between two straight lines which are parallel (as their only point of intersection is inverted to infinity). The geometry of such a set up is easy, as all the circles are congruent, and the problem reduces to, for each circle, finding the radius (which is constant) and the distance from the center of the circle to the center of the circle we used for our inversion. This data directly determines the distance of the nearest and furthest points on the circle from the center of our inverting circle, which determines the corresponding data when we invert back, allowing us to calculate the radius.

For a worked example using this method, see this video.

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